1.

The boiling point of water is 100^(@)C and it becomes 100.52^(@)C if 3 g of a non-voltile is dissolved in 20 mL of it. Calculatethe molecular weight of the solute. (K_(b) for water is 0.52 K kg mol^(-1), density of water= 1 g mol^(-1)).

Answer»


Solution :`DeltaT_(b)=100.52-100=0.52^(@)C=0.52 K`
`W_(B)= 3 g, W_(A)=20 g [20" mL of water= 20 g"], 0.02 kg, k_(b)=0.52" K kg mol"^(-1)`
`M_(B)= ?`
`M_(B)=(K_(b)xxW_(b))/(DeltaT_(b)xxW_(A))=((0.52" K kg mol"^(-1))xx(3G))/((0.52K)xx(0.02 kg))=150" g mol"^(-1)`.


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